text:
Why does the vortex created by wing affects its own angle of attack?
According to potential flow theory, we know that the lift generated by the wings is due to the vortex it creates. In case of a 3D finite wing, there is also a span wise flow due to tip leakage from bottom to upper surface. So if suppose the wing is moving towards left, then the direction of vortex should be clockwise, which in turn should produce an upwash rather than downwash. This should cause an increase in angle of attack rather than decrease.
So the downwash generated should be valid for a wing travelling behind the original wing, but we say that there is downwash on the original wing.
So where is it that I am making a mistake, vortex direction. If yes, what is the correct explanation. Please explain.

The wingtip vortices create both upwash and downwash; The downwash lies within the wingspan and affects the wing angle of attack while the upwash region lies outside the wingspan and can be utilized by another aircraft (or bird) flying behind and above the wing.
The flow induced by the downwash reduces the effective angle of attack of the (finite) wing and causes the induced drag.
See if this image clarifies your doubt.

Source: aerospaceweb.org

Consider the upwash before the wing and downwash after the wing (section).

Source: theairlinepilots.com
As it can be seen, the airflow in front of the wing is slightly directed upwards by the upwash. Now, the angle the wing makes with the horizontal datum (i.e. relative airflow) is the same.
However, the airflow is slightly turned upwards and the wing sees the air coming at it at a different angle compared to the relative airflow. As a result, the effective angle of attack (defined as the angle of attack that lies between the chord of an airfoil and the effective airflow) is reduced, so that the relative and effective airflow looks something like this:

Source: theairlinepilots.com
This difference between the relative and effective airflows is what that causes the induced drag.

Source: theairlinepilots.com

Perhaps you could also mentiond that the same also happens in front and after the wing. Directly in front of the wing upwash occurs, wheras behind the wing downwahs occurs, see https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcQ1ggS-L23tSS1YyVJyfBUNkX6Y1kRacLVnvjxEfiU5z5n3z9ZQgg

@aeroalias Thanks for this very relevant image. But that exactly is my point, if all the induced flow effects are prevalent behind the wings, how is it so that the angle of attack which depends on flow ahead of the wing is getting altered?

@ROIMaison if upwash occurs ahead of wing, then shouldn't AoA increase?

@Manish See the edited answer.

Thank you very much for the detailed explanation, starting from the fundamentals. I think i was confused with vector addition of velocities, whereas in reality, the flow is getting rotated.

Why is there an up wash. The vortices only start on top of the wing so why would the vortices cause up wash ahead of the wing

How is the angle of attack reduced if the flow is pushed up. The angle of attack would increase.

@Crafterguy and aeroalias: part of the answer is not true! The wingtip vortex produces a downward velocity component ahead of the wing, and also behind it. See my answer for more info.

Near the wing the bound circulation due to lift leads to an up-wash ahead of the wing and downwash behind the wing similar to the flow produced by a two-dimensional lifting wing of infinite span.
A very important effect is generated by the flow due to the vortex pair that comprises the wake (at wing tips). The semi-infinite sheet of vorticity distributed in the wake produces a downward velocity component in the free-stream ahead of the wing, at the wing and far downstream as illustrated in Figure.

The downwash by the wake leads to a reduction in the angle-of-attack of the wing
relative to the free stream, reducing the lift. In addition the downwash rotates the oncoming flow vector at the wing leading to a component of drag as shown. The change in angle of attack due to the downwash generated by the wake is tg(a) = (Uz/U∞)
Source:
STANFORD UNIVERSITY,
Department of Aeronautics and Astronautics,
AA 200A Applied Aerodynamics,
Instructor: Brian Cantwell, cantwell@stanford.edu
Link:
AA200_Ch_12_Wings_of_Finite_Span_Cantwell.pdf
Web folder containing other stuff: URL


text:
Is it possible to determine the mass flow rate through the engine of MiG-21 during its various cruise Mach no. operations?
Is it possible to obtain the mass flow through the Tumansky R-25 turbojet engine mounted on a MiG-21 during its various standard operating (cruise) Mach numbers?
Note: standard as in older planes where there were just 3 positions of the inlet cone - upto Mach 1.5, between 1.5-1.9 and above 1.9.
Also, what should be the cruise condition for this plane, given that it is capable of flying at various Mach numbers and thus speeds?

Just out of curiosity, what is this project you are working on?

oh it is a homework project :P I suppose assignment would be more appropriate term.

I wish my homework had been like that...would have probably helped if I'd been doing aeronautical engineering of course...

Tell me what is given and I can start to answer. Otherwise, just google it ;-)

I don't have any data on the Tumanski engine, only on comparable western fighter jets of the same generation.

General Electric J79 mounted in the F-104 Starfighter: mass flow = 77 kg/s.
Snecma ATAR9K50 mounted in the Mirage III: mass flow = 73.5 kg/s.

These are the max. mass flow rates, at maximum thrust. They don't scale much with the thrust at operating RPM, usually the inlet nozzle is choked.

source: Stack Exchange

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sort: votes


text:
What are the pros and cons of a straight wing plane?
I need this asap for my folio... Much appreciated. I'm doing a research section on my folio and I need to know the pros and cons for a straight wing in general. Everywhere I've searched doesn't seem to have the question above and is making it very difficult.

I assume you mean unswept vs swept-back wings.
Pro:

Easy to manufacture
More efficient at low-subsonic speeds

Con:

Unstable at transsonic and supersonic speeds (see Mach tuck)

More details about wing sweep (including forward sweep) that should answer your question: Swept wing

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text:
What the weather condition can affect for flight?
What is the most disturbing weather conditions for flight?
On another word: what more dangerous for takeoff/landing: wind, precipitation or something else?

I am working with a prediction model for delay regular flights and wondering what can be more important for this.

It's actually fairly simple when talking about weather related dispatch delays for the airlines.
Number one items:
In summer, thunderstorms, especially those associated with fronts and squall lines.
In winter, snowstorms with heavy accumulations, or severe freezing rain.
And beyond that, in both summer and winter, extreme low visibility, as in fog.


text:
How come Boeing invested \$32 billion on the 787 program and Airbus only \$15 billion on the A350?
How come Boeing invested \$32 billion on the 787 program and Airbus only \$15 billion on the A350? (according to wikipedia)

I added a backslash before each $, otherwise the forum software thinks you're using MathJax.

Here are some reasons you could find in Wikipedia English and or French versions which are not exactly similar
The B787 composite body study was very expensive, indeed Airbus built its fuselage more conventionally from four large panels, two forming the flanks, to which are added the lower panel and the flag while that of the B787 is in one piece in autoclaves. Being in one piece the study of each models 787/8/9/10 required a separate Study.
Boeing had repetitively fuel leak problems requiring deep extra studies while the production was already launched.
The 787 APU had unexpected problem, mainly on ground, in some circumstances after being switched off it couldn’t be restarted before a delay of 2 hours.
You know the decision of the FAA to ground the aircraft for more than 3 months due to batteries risk of fire
The A350 is only 50% produced by suppliers outside the group, compared to 70% for the dreamliner, this is a source of coordination and logistic problems
Aware of the lithium batteries problems, Airbus did initially a conventional choice, thus avoiding delivery delays.


text:
What is the average time and distance needed to reach cruising height?
I am trying to compare trains with airplanes and the distances where taking a train is more viable than an airplane. I guess this would depend on the type of plane, but the Airbus 320 seems like a common enough plane in Europe so I have something along that type in mind.
As far as I know, it would be a good model for airplanes to separate the takeoff and landing phase, where they cover less ground and does take appreciable time on shorter hauls and the cruising phase. I found average speed for the latter but I don't know how much time takeoff takes and how much ground the aircraft can cover during this. Same goes for landing, so my question is, how much time do these take and how far does the aircraft get during these?
I'm also assuming that when looking at city center to city center travel times, taking a plane has about a three hour overhead (airports are far and it takes quite a long time to get in, security check and all), but if anybody actually knows some statistics on this I'd be much obliged :)

There is a simple rule of thumb in Europe. For domestic trips, trains always make more sense. For international trips, airplanes make more sense. The reasons are time, pricing and heavy competition in international vs domestic routes. Airliner performance does not play into this.

I am right now only interested in time, nothing else. Also, your rule of thumb might apply for Spain, but Vienna - Budapest is already faster and cheaper with a train and there's nothing really fancy going between the two :)

"Takeoff" and "landing" have rather specific meanings in aviation, and are very short in duration. You are probably more interested in the climb and descent portions of the flight.

When flying, nearly no time is spent below average ground speed of even a high-speed train, so in reality flying speed isn’t the controlling parameter, I believe... nonetheless, good question!

I think this question needs to be restated, the title asks a specific question but the post's body seems confused.

@fbence What makes you think I am talking about Spain? Vienna to Budapest is indeed the exception. And about the only exception. If you are only interested in time, it takes 45 min by air, 2.5 hours by train, so if you have time to waste... :)

@JuanJimenez Is that rule from your left thumb or your right thumb?

@koyovis Still confused, eh?

@JuanJimenez Spain: your name and that I just spent a lot of time taking trains all around Spain :D Vienna - Budapest: well, it takes 2.5 hours by train, while it takes almost 2.5 hours to get on a plane, so definitely a lot faster by train :D

@aCVn Sorry, yeah the descent and climb is what I'm interested in because it seems to me that it is a slower part than cruising and on short flights I thought it might be a pretty relevant part.

@CptReynolds Yes, I guess it's faster almost always, than a train, but my thinking was that on shorter trips the climb/descent, which seems to me to be slower than cruising, might make up a large portion of the flight and so didn't want to cheat in favor of planes by taking the average cruising speed for the entire distance.

Yes that’s good thinking. On most aircraft and flights, it’s within 5 minutes after takeoff that 250kts (indicated airspeed) is reached, which is higher in true airspeed depending on altitude, but could be even lower in ground speed due to wind influence. Up to 10000ft, most flights are limited to 250kts. Depending on aircraft type, above that, something around 280-300kts indicated airspeed is flown up to the altitude where that speed equates the Mach number flown from that altitude to cruise altitude (something in the Ma .72 to .82 range will probably work). Descent is similar in reverse.

@fbence You got it wrong by 4000 miles.

I answer directly to the title of the question:

What is the average time and distance needed to reach cruising height?

The Takeoff takes very little time, but the climb phase depends on the cruise flight level clearance. Assuming you climb with a vario of 2000ft/min with an average IAS of 200kts. The time to reach the cruise level is equal to:
(Cruise level in ft - takeoff terrain elevation if ft)/2000, the result is in minutes.
So if the departure is at sea level and the cruise altitude is at 30000ft, the climb phase will take : 30000/2000=15min
The descent till arrival could be more complicated because in addition to the pure descent you have to level before intercepting the approach phase which occurs relatively at lower speed. If you are lucky, not having to hold above the airport, the total descent might take 20 to 30 minutes
Routes are not necessarily straight lines, nevertheless you may assume an average cruise True speed of 500 knots.

Thanks for answering, but I don't think I actually understand your answer, which is probably due to my lack of knowledge in aviation. I'm guessing that the 500 kts/hour would be true for cruising. You also said climbing and descending will take about say 40 minutes in total. I'm assuming during this 40 minutes the average speed is below 500 kts/hour, so what would be a good estimate for the amount of ground covered in that 40 minutes?

After surfaces retraction the speed goes rapidly to 250 kts indicated, but at sea level no real difference between true speed and indicated airspeed. However at cruise altitude even though being at about 250knots indicated, your true speed may be about 500 knots. So you may assume during climb and descent  phases an average true speed of 375 knots. You are right, these phases should be calculated separately, I do apologize.  For very long flights roughly we can consider an average true speed of 500 knots, I will correct.

"500 kts/hr" would be a (weirdly expressed) rate of accelleration, as knots is already nautical miles per hour, thus kts/hr would be nm/h/h.

Yes indeed, I wanted to write miles/hour instead of kts, to be clearer ....unhappily I did a lapsus

I thought it  was addressed to me, I am sure fbence did a lapsus

yeah, I messed up the knots/hour :) @user40476, if you add the correction to the answer I will except, thank you!


text:
What does the “test result indicator” indicate on a cockpit voice recorder?
I am really curious as to what the “green/red” strip of led indicate and the numbers below it. Having read around the web, i speculated that it could be an audio quality rating scale. Would help alot of someone could please help explain the “numbers” and why there is so many led blocks on the strip...

There are a number of different CVR models out there, which one is this specifically?

Looks like a newer Airbus CVR panel. The test button generates an internal low frequency test tone signal and the LED will tell you the quality of the audio picked up by the CVR when the tone is generated. The tone is internal and can only be heard if you plug in a headset into the headset jack. A good test will indicate in the green, the more green, the better the test result. On the A320 the tone is heard over the cockpit speakers.

@RonBeyer i am not too sure about the model, but its from A330-300

@JuanJimenez By any chance do you know the meaning of the “numbers” below the strip? 0.3-1.15? So could you say, a new plane (if tested on the CVR) would get a full green led on the strip?

I have not been able to find any other example of a panel with the numbers, let alone an explanation. Sorry. I am also not 100% sure of the details, which is why I posted as a comment and not an answer. But if I had to guess, the numbers may be referring to the measurement of the frequency of the tone and how it compares to what the ideal measurement should be. But that is only an educated guess.

I could not find the exact model you showed, but the A330 and A340 FCOM on SmartCockpit shows a similar one. Unfortunately the description of the test is rather short:

The Overhead Pushbutton explanation on SmartCockpit is equally short:

My best guess would be that it shows the recorded volume from low on the left to high on the right and it is considered OK (test passed) if it reaches into the green LEDs, therefore making sure the CVR can pick up voices clearly enough. I hope someone working with these CVRs can provide a more accurate answer!


text:
Do aeroplanes have to balance torque like helicopters?
Helicopters need a tail rotor to balance the torque reaction that tends to rotate the main body sideways. Doesnt a single-engine plane suffer the same problem? I would imagine that (for example) the left wing would dip lower than the right due to the torque. How is it countered?

Not only torque reaction. Yawing and pitching moments,  asymmetrical thrust and gyroscopic precession as described here: Effect of Propeller on Airplane Dynamics

A single engine plane can suffer from control problems due to torque.
An early famous example was the Sopwith Camel
As others have pointed out, torque effects have to be countered using the normal aerodynamic controls available to the pilot. Some aircraft had extra-large ailerons because of this issue.

It should be mentioned that the rotary engines cause much bigger problems with precession than other types making Sopwith Camel one of the really bad cases.

Indeed, what we are seeing with the Sopwith Camel is not a simple case of propellor torque. The entire engine is rotating about a stationary crankshaft (for cooling reasons) and the acceleration of the engine is largely responsible for the rocking. According to the video commentary, the engine has no throttle and only 3 power settings (3,6,or 9 cylinders) so the pulsing is actually a design feature! It must have been vary waring to fly. Precession refers to the fact that the engine also acts like a massive gyroscope. This plane could turn much faster and tighter in one direction than the other

In fact, fighter pilots of the time were very aware that, due to those forces, planes could turn in one direction (the "rotary turn") much more tightly and faster than in the other. Important to know when dogfighting!

In contrast to what happens in helicopters, the static and dynamic stability will prevent big movements of propeller aircraft due to engine and propeller torque. The biggest consequence, the rolling moment, is taken care of by roll damping. When the propeller torque rolls the aircraft, the downwards-moving wing will see a higher, and the upwards-moving wing a lower angle of attack. Both wings will create a rolling moment which will quickly stop the rolling motion. All the pilot has to do is to restore the correct roll attitude using the ailerons.
The lift difference of the rolling aircraft also creates different drag on the left and right wings, so a small yawing moment will turn the aircraft slightly sideways when the rolling starts. This has to be corrected by rudder input.
If we look at other secondary effects, the yawing motion will cause a slight pitch change by way of engine and propeller precession. This also will be stopped quickly by pitch damping (the pitch movement changes the angle of attack on the horizontal tail surfaces), and the pilot needs to correct the pitch attitude again.
In consequence, changing RPM will induce slight movements around all three axes in a single-engined propeller airplane.

It is countered by the natural stability and the roll control provided by the ailerons. If you have insufficient airspeed for roll control then you will be on the ground and your gear will prevent excessive rolling.

There is a big difference between the amount of torque delivered to the main rotor of a helicopter and that delivered to the propeller of a comparably-sized fixed wing aircraft. Also, without the tail rotor, a helicopter in a hover has nothing to "push against" in order to deliver the needed torque to the main rotor. If not for the tail rotor, torque delivered to the main rotor would cause the rest of the vehicle to spin up in the opposite direction. Hovering flight would be pretty much impossible without something to aerodynamically generate opposing torque (tail rotor, second counter-rotating rotor, etc.).
Torque delivered to the propeller of a fixed-wing aircraft is opposed either through the ground (when stationary, taxiing or on take-off roll), or by the wings (with minor roll trim or control input) when in flight. As explained in the other answers, propeller torque does induce certain effects, but they are of much less significance than that of a helicopter main rotor, and are dealt with by trim and pilot awareness/handling.

Torque is more noticeable in aircraft with tail wheels (tail draggers) which also have large engines (e.g. world war 2 vintage) for two reasons:

If the throttle is advanced too quickly, you might not have enough airflow over the rudder to be able to counteract it and a swing followed by a ground loop could result.
At the point when the tail is lifted off the ground during the take off run, gyroscopic effects occur that could cause a swing (try holding a spinning top and then move it around - it exerts forces in unexpected directions).

Also on aircraft with the largest engines, the torque could make one wing lower than the other during the take off run. You also have the problem going around to be very careful with both throttle and rudder because these effects are more pronounced at lower airspeeds where airflow is less.
An aeroplane with a nosewheel is generally more stable on the ground (due to the CG being ahead of the main wheels rather than behind), so these effects are less noticeable.

Sorry yes you are quite right, p factor is another thing in addition to the gyroscopic effects I mentioned. Searching for "p factor aviation" gives a quick explanation from Google

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text:
What is the length of the escape rope in the cockpit? specifically A330-300 (If possible)
I understand you could measure from height of the cockpit window, but is there any official dimensions stated anywhere such as FCOM or FCTM documents?

Do they suppy gloves as well for rope burn?

@securitydude5 If you are escaping the cockpit of a burning aircraft I'd be grateful for the rope instead of the broken legs, burned hands would be the least of my concerns... I doubt that this info is in the FCOM or FCTM. This may be in a maintenance manual somewhere or specified with a part number (ex: 00001ABC 6M Escape Rope). I'm not even sure that this would appear on a MEL.

@RonBeyer I agree with you fully but for heavier guys, I doubt there'd be skin left on their hands by th time they get down they might even let go mid way and break their legs from the pain&as for the ladies, I do fear for them however thieir size.  Doesn't have to gloves with 5 fingers cause wearing those may take time, it could be a square or rectangular material designed for that purpose, so easily accessible that if you have time to let down the rope, you should have time to pick it up.

It could be the bag the rope is kept in or incorporated into the rope, always at the top end of the rope.  The FCC should consider making it compulsory.  Engineers could come up with a better design rope tha incorporates safety, even the risk of falling of and braking many bones because some people may not have the strength or master the skill of dis embarking off a rope worse still under threat of a fire.  Consider over 55 pilots flying a boeing 747 or an Airbus A380

To be on the safe side, it's always best to bring your own rope.

For the A320, this can be found in the CCOM (Cabin Crew Operation Manual)...

So the A320 is a 5.5 meter (17.04 foot) knotted rope stored above the side cockpit windows. There are 2 ropes in the cockpit (one on each side).
I can't find the CCOM for the A330 but it should be a similar size.

(From A330 CCOM) The length of the rope is 6.7 M. The maximum load is 181 Kg (400lb).

The length makes sense, but do you know why the A320's rope can support 900kg (see Ron Beyer's answer), but the A330's only 181kg?


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Required RVR for Landing
There are a few questions out there on this topic, but I have some confusion about the wording and the meaning of RVR and its usage,
"The touch-down zone RVR is always controlling. If reported and relevant, the mid point and stop end RVR are also controlling. The minimum RVR value for the mid-point is 125 m or the RVR required for the touch-down zone if less, and 75 m for the stop-end. For aeroplanes equipped with a roll-out guidance or control system, the minimum RVR value for the mid-point is 75 m.
Note: “Relevant”, in this context, means that part of the runway used during the high speed phase of the landing down to a speed of approximately 60 knots."

"...Or the RVR required for touch-down zone if less..." Does this that if the RVR is less than 125m, use the touch down zone RVR?

The word "Relevant", does this mean that if the plane will be over 60kts at the mid point, then the RVR minima here is to be considered? but if plans to be less than 60kts at the end point, no need to consider the minima there?

What is the touchdown zone RVR? would this be what is on the approach charts?

Thank you!

Welcome to Av.SE. It would be helpful if you can provide a link to the document that you are quoting, because context matters when working to explain a particular statement. Thanks!

Thank you! The document is actually from the company's OM-A, but I think it must have been an extract from an ICAO annex. But maybe just to confirm my final understanding from a high level perspective, its essentially trying to make sure you apply the right RVR, based on your approach type and landing distance required?

If you are performing a CAT3 without DH where RVR required for touchdown zone is less than 125m, the requirement for the middle zone is the same as it is for the touchdown zone. Note: you have to perform an automatic rollout.

Yes, but consider them as zones. So if you have calculated the required landing distance of about 2km, the last kilometer of a three kilometer runway is not relevant.

Assuming you have the required equipment, training and all the ground facilities including the lights are operational, then yes.

As a sidenote to question three, for every approach you are flying from visual approaches to low visibility operations the required RVR for approach is always required only to be available in the touchdown zone. For middle and stop ends, 125m and 75m applies.


text:
Migration from AFTN to AMHS?
My question is:
Is there any obligation from ICAO about migration from AFTN to AMHS? and If that's so, what are deadlines? Any ICAO doc about this?


text:
Final 200 feet under autoland
How does Autoland work below decision height (200 ft)? Ie, what guides the plane until the radar altimeter notes the jet is at the right height to begin to flare? I've been searching the internet, and it looks like the plane simply holds its attitude for around 100 feet until the RA kicks in.

An autoland can be performed when a CAT II or III ILS (Instrument Landing System) approach is available (bad things can happen otherwise: Why don't pilots always use autoland?). The decision height for these approaches is lower than the usual 200 ft:

Cat I: DA 200 ft or higher
Cat II: DA 100-200 ft
Cat III A: DA < 100 ft
Cat III B: DA < 50 ft

The pilots must have the runway (or runway lighting) in sight when reaching the decision height. Otherwise, a go around must be executed. But this does not mean that the ILS signal is no longer available. The autoland system continues to use the ILS signal together with the RA (radio altimeter) to fly the aircraft. That means the autopilot uses the localizer deviation to control the aircraft laterally (during the approach, the flare and even after landing) and the glide slope deviation vertically until the RA indicates it is time to start the flare (at this time glide slope is ignored).
The exact details will of course depend on the aircraft model. Here is how the autoland works on a Boeing 737 NG (source: FCOM v2 4.20.19 Automatic Flight - System Description):

500 Feet Radio Altitude
The pilot is required to check for the presence of LAND 3 or LAND 2 in order to
  continue the autoland.
  If the second autopilot in CMD remains armed and does not engage, LAND 2 or
  LAND 3 does not annunciate. Instead, the amber NO AUTOLAND annunciation
  alerts the pilot that dual control has not been established and the autoland is to be
  discontinued.
450 Feet Radio Altitude
The alignment mode is enabled which provides rudder compensation for the
  purpose of decreasing large crab angles produced by crosswinds, and to control
  the adverse moments caused by an engine failure. The automatic correction for
  aircraft crab angle due to crosswinds and engine failure enhances flight crew
  runway perspective and provides optimal aircraft position for initiation of rollout
  control. In a strong crosswind, the airplane does not fully align with the runway
  but lands in a slight crab. Sideslip is limited to 5 degrees. This mode is not
  annunciated.
400 Feet Radio Altitude
The stabilizer is automatically trimmed an additional amount nose up. If the A/Ps
  subsequently disengage, forward control column force may be required to hold the
  desired pitch attitude.
If FLARE is not armed by approximately 350 feet RA, both A/Ps automatically
  disengage.
Flare
The A/P flare maneuver starts at approximately 50 feet RA and is completed at
  touchdown:

FLARE engaged is annunciated and F/D command bars retract.
FLARE engaged is annunciated and with LAND 3 annunciated, F/D
  command bars center.
the A/T begins retarding thrust at approximately 27 feet RA so as to reach
  idle at touchdown. A/T FMA annunciates RETARD.
the A/T automatically disengages approximately 2 seconds after
  touchdown.
the A/P must be manually disengaged after touchdown. Landing rollout is
  executed manually after disengaging the A/P.

Rollout
ROLLOUT arms when LAND 2 or LAND 3 annunciates.
  At approximately two feet radio altitude, rollout activates:

ROLLOUT replaces the VOR/LOC roll flight mode annunciation
the autopilot controls the rudder and nose wheel steering to keep the
  airplane on the localizer centerline.
rollout guidance continues until a full stop or until the autopilots are
  disengaged.

This is a good answer but does not address the particular guidance and control aspect of the question (yet?).

@CptReynolds I added some more details. Not sure if that answers all questions.

Autoland can only be performed with CAT II/III as far as I know. The minima of CAT II/III are usually quite a bit lower than 200ft AGL. Also, as the localizer is placed at the far end of the runway, the aircraft will not overfly it and therefore not lose it on landing. The aircraft will just follow the localizer and glideslope until the radar altimeter triggers the flare and leaves the glideslope.
To improve the reliability (or reduce risk of disturbance) there is also a second hold-short line a bit further away from the runway which is being used when CAT II/III is in progress. This assures no other aircraft reflects or blocks localizer or glideslope signals.

I'm not 100% sure bit I even think CAT IIIb is required for autoland which means decision height below 50ft.

Autoland can be performed under any approach category, but only with Cat2/3 there is assured signal protection. A Cat 1 autoland can be performed, but crew will be very aware of the possibility of having to take over.

And indeed there will be a lowest altitude after which glideslope signal will not be considered, depending on aircraft type even before the flare, because the „funnel“ is too narrow for effective guidance. A sensible sub-mode for guidance e.g. between 80 ft and flare could be to hold vertical speed or fly nominal path angle or so. I can look it up in the next days if not answered before.

The last comment  is exactly where I was heading. From my research, when autoland was first invented the biggest problem was guiding the plane's downward trajectory under 200 feet because glideslope became unreliable at that point. My readings say the plane was allowed to fly "ballistically," ie, maintaining attitude so that the trajectory holds until the RA determined the flare should start. I am curious if in the past 50 years anything has improved, ie, is glideslope accuracy good now under 200 feet? Cpt Reynolds, are you saying glideslope is good to 80 feet? So 30 feet of "ballistic flight?


text:
About F-35 hover/ roll posts responsiveness
how responsive is the f35 in hover. especially in the roll axis.
What do think of the roll posts? is it a good concept?

The pilot controls have no direct connection to the attitudes (like roll) in hover mode.
Instead the throttle and stick is used to tell the computer if you want to go up, down, forward, back, left or right.
So for example pushing stick to left by some degree tell the computer to move left with a certain speed, the computer will handle all the roll for that. So pilot controls sideways speed, not roll.
Btw. this is probable relevant to your question: How does the F-35's roll posts work, and how does its engine turn down 90 degrees?

thx for the answer. but perhaps i should have been more specific. I wanted to know effective the roll posts are in providing stability(whether the pilot or the computer controls them). do you think roll posts would be more effective than control surfaces in flight?(for stability and maneuveurability).

Ah yes, when you asked "how responsive" I thought you meant pilot control input to roll it. I do not know the answer nor have an opinion on your question then.


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Which part of the boundary layer(chordwise) contributes most to friction drag? fore? aft? Why?
along the chord of a wing/airfoil section, which portion of it creates the most friction drag through the boundary layer?
Is it in the fore section, close to the stagnation point? or the aft section

It depends on the airfoil and the free stream conditions, but some general trends can be observed.
First, skin friction drag is dominated by viscous effects, so the laminar or turbulent quality of the layer is important. Laminar layers are less draggy that turbulent ones, mostly by merit of being thinner. This means that the leading edge will often suffer less viscous drag until the boundary layer transitions from laminar to turbulent, which may happen next to the stagnation point or further downstream. The layer will then continue to grow, so, for attached flow without re-circulation bubbles, the friction maximum can be expected to be near the trailing edge (not necessarily on it).
Second, for this viscous friction to be drag in the usual sense, it has to provide a force opposed to the direction of flight, so one would integrate its product with such a vector along the surface of the airfoil. This would mean, again, that the leading edge contributes less than its fair share, as friction forces directed up or down relative to the free airstream are not drag.
So the general answer would be that the aft part of an airfoil provides a larger friction drag contribution than the fore.


text:
Are subsonic nozzle fully expanded?
this is about converging subsonic nozzles.
Are they fully expanded? meaning exit pressure=ambient pressure?
What about diverging subsonic nozzle?
Are they fully expanded? what would happen if the exit pressure was greater than ambient? shock formations?

Fully expanding nozzles in the subsonic regime... wouldn't that mean exit velocity = 0? This would be great for a turbo-prop engine but not for a turbofan/turbojet engine. Thrust on turbofan engines is often measured as an engine pressure ratio (EPR). During takeoff this ratio can reach 1.5 or 1.6, which means the exit pressure is 1.6x greater than the intake pressure. Shocks only form when the air exits at super-sonic speeds.

Yes, to convert the built up pressure and utilize its full potental, they are typically fully expanded.

In some cases it might be good to overexpand your exhaust, to keep it behind the engine

@Jan The static pressure would be 0, and the dynamic pressure (velocity) would be maxed out.
Static pressure + Dynamic pressure = Total pressure.
After combustion Static pressure would be high and dynamic pressure would be low.
eg. 10 Sp + 2 Dp = 12 Tp
Moving down the converging duct, the static pressure will gradually fall, and the dynamic will rise.
About halfway.
6 Sp + 6 Dp = 12 Tp
At the end of the converging duct (if we have full expansion, all static pressure will have been converted to dynamic pressure - velocity).
0 Sp + 12 Dp = 12 Tp


text:
what is the physical explanation for additive drag?
from what i understand, additive drag is not really drag. it is a way to account for the variation of inlet condition due to conservation of mass.
 but i still don't get why it create drag(negative thrust) when bookkeeping the thrust value of all parts of jet engine/ducted fan
can anyone explain this for me

Related, possibly duplicate: https://aviation.stackexchange.com/q/29376/17780

The question is perfectly clear. Why do people flag it as unclear?


text:
Is adverse yaw ever desirable?
Is there any situation during flight where adverse yaw (yaw due to roll) is desirable?

Yes. For deliberate cross control situations, adverse yaw helps.  This means during side slip or forward slip maneuvers, where adverse yaw means less rudder is required to keep the nose pointed in the opposite direction to the lowered wing.
If you are in a glider or an old classic aircraft with lots of natural adverse yaw, if you lower a wing without any rudder to keep the ball centered, you end up in a mild side slipping turn automatically thanks to adverse yaw.


text:
Why are some exhaust cone exposed?
Some engines have an exposed tail/exhaust cone at their exit, protruding aft the nozzle.
how does that affect the overall thrust and pressure distribution along the walls of the cone?
is it's role similar to an aerospike?

I think an example could be the A320 CFM 56 vs. IAE V2500 engines. The nacelles on the IAE V2500 are longer, which is based on the design decision to mix the bypass airflow and core engine mass flow inside of the nacelle area which as far as I know can increase the propulsion efficiency but costs extra weight. On the CFM the bypass air is let to the outside right away which saves weight on the nacelles which decreases propulsion efficiency but also decreases the drag of the nacelles and thus can increase the overall efficiency. Just a design choice AFAIK.

Related: https://aviation.stackexchange.com/questions/37959/how-do-exhaust-cones-affect-the-performance-of-turbine-engines?r=SearchResults and https://aviation.stackexchange.com/q/37938/4108

Also https://aviation.stackexchange.com/q/20943/4108

thx but ive seen those threads already. still doesnt answer my question


text:
How to calculate pressure distribution over an airfoil from its coordinates?
I need to compute the pressure distribution over an airfoil (I have the coordinates, but any will do) with the streamline curvature method. Unfortunately, I cannot find much on the web.
Can someone point me in the right direction or walk me through it? I understand it qualitatively (the concepts) but have a hard time quantifying it (actually solving the flow field).

Another detailed source can be found in this paper by Mark Drela on xfoil (https://web.mit.edu/drela/Public/papers/xfoil_sv.pdf). For your purpose, you can stop at section 2.1, assuming source strengths at every panel to be 0.

What helped me was the book "Theoretical and Computational Aerodynamics" by Jack Moran (ISBN 0-471-87491-4). That was 30 years ago, but the physics haven't changed since then. And it is still available.
An online source would not go deep enough to cover all what you need. My advice is to get the book and work through the exercises. It worked for me, and if even I could code a 2D potential code after having read this book, you should have no problems.
There is one trick: At one point you will end up with n equations for n+1 unknowns. Bummer! What you need to do is to prescribe flow direction at the trailing edge, and then you will have n+1 matching conditions for those n+1 unknowns, n being the number of panels you divide the airfoil surface into. Google "Kutta condition" for an explanation.

EDIT
Now you have quite a different question:

by CFD i meant RANS. how does it compare to RANS methods in term of stall prediction and pressure distribution

@toshiba: The comparison is like between a Volkswagen Beetle and a Mercedes. A potential code is simple, robust and gets you there, like the Beetle. Pressure distribution is good until separation starts. Compressibility can be modelled until locally the speed of sound is reached. Stall prediction is a bit tricky, but with a bit of experience and an added boundary layer calculation you will get good results.
RANS is a simplified Navier-Stokes algorithm which is used for modelling turbulent flows. Like the Mercedes, it needs much more effort to code and run (and computing power - it requires a mesh where the potential code only needs panels on the airfoil surface) but also produces better predictions if the mesh is laid out well. Drag and stall approximation is included while potential flow needs some extra effort for boundary layer modelling. A Navier-Stokes code can also be used for transsonic flow.
If you want to build your own, assume you get the parts only and have to assemble the cars. No plans included. The VW Beetle heap is much smaller and you can figure out quite quickly how it fits together (did that once with a Beetle engine that someone else had disassembled and left behind). When you are finished, the car will run and require little maintenance.
With the Mercedes, there are many more parts, more complicated looking and requiring special tooling. It takes much longer to get assembled and in the end you will have some parts left over and others in the wrong place if nobody experienced guides you along. Maybe the car runs, but it will break down quickly unless you put much effort into maintaining it.
In other words: Start with a potential code. MUCH easier.

will this allow me to compute the pressure distribution over the entire airfoil only, or the entire flow field as well? Also, how accurate is this method vs experiments or CFD?

@toshiba: The method uses potential theory, so you get a solution for every point in the flow field. It is the boundary conditions which have to be met on the airfoil surface. Potential theory only works well for attached flow and is inviscid. However, Moran also includes a simple boundary layer calculation so you will also get an approximation of drag. And it is CFD (but a rather simple variety of it).

great thx. by CFD i meant RANS. how does it compare to RANS methods in term of stall prediction and pressure distribution. Can it successfully model compressibility effect? Im sorry about the question but i need to write a code and i chose potential flow theory beacause i was under the impression that accuracy would be sufficient. but im having doubts now. how good is that drag approximation?

@toshiba: That is a new question altogether. See the edit.

thx. however i must add: i just read the book you recommended. It is based on potential flow yes, but not streamline curvature method.


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What is this cylinder covered in threaded studs and what is it for?
I can't figure out what this is.  It looks like each post is threaded, but why is a mystery to me.  My guesses include a grounding post, a bolt/nut testing stand that's not actually in an aircraft, or a super metal black box .
What is it and what does it do?

good question.  it was in a collection of images of airframe components from an inspection, so i assumed it was.  i'll try to go back to the source and verify.

You guys are jumping to conclusions because you don't know what you are looking at.  Answered.

It's the inner liner of a heat muff, either for cabin or carb heat (you can see the engine mount tubes right next to it and that's exactly where I'd expect to see it, and the black pucks behind are part of a nose gear shock absorber).  It could be home-made, using threaded rod welded to the inner liner, or maybe it's a manufactured component with the rods swaged in place like rivets.
An outer shroud will go around it with inlet and outlet bosses for SCAT hose running to the air inlet and outlet to the cabin or carb.  Sometimes springs are wound around, or some other metal that can act like cooling fins; some builders just stuff the space with stainless steel wool.
The threaded rods are to improve heat transfer from the inner shroud to the air flowing past them (the threads maximize surface area).  There will be a flange out of view at each end that creates an enclosed annular chamber.  The inner shroud itself typically has internal baffling to function as a muffler for the exhaust.  Crude sketch of what I mean below.  My own plane has this kind of thing, but without the studs, just an open space.

"the threads maximize surface area" Exactly. Nowadays I'd assume fins to be used instead, but bolts used to work just fine. This is also why the rows are shifted half a column each time, to increase heat dissipation.

Great work! And thanks for the visual too!


text:
Why do wingtip vortices affect the upwash ahead of the aerofoil?
I understand that wingtip vortices increase the downwash aft of the aerofoil and the theory that they will alter the effective airflow angle, reducing effective angle-of-attack, with the total reaction of lift acting at 90 degrees to Effective airflow, not the relative airflow which now has an extra component of induced drag.
But why does this happen? How do the vortices increase the upwash ahead of the airfoil when they occur rearwards of the wing?

In subsonic flow, is it not expected that any perturbation of the airflow will affect the entire flowfield? The magnitude of this effect can of course be very small.

This is a fascinating topic, and it really boils down to the mechanics of lift. Physically, upwash and downwash are generated as the air accelerates/decelerates and curves around the LE of the wing. At the wingtips, higher pressure air curves upward and generate the vortices we are familiar with. In fact, there are vortices generated at the trailing edge of the wing, too, just not as noticeably. Obviously, this picture is unsatisfying since it doesn't give us any insight to lift.
According to the potential flow theory (irrotational, incompressible and inviscid), lift is generated whenever there is a net circulation of air. A particularly successful model, the Lifting Line Theory, models the span of the wing as a collection of U-shaped horseshoe vortices: there's a bound vortex at the span location, and two trailing vortex arms extending to the far field.

So there you have it. A tiny downwash is generated by an individual vortex as it turns the flow downwards at that location, and a tiny upwash is generated as it turns the flow upwards. If you proceed to do the math and sum over all the influences at any point on the span, the net effect is a downwash (which changes the effective AOA and gives rise to induced drag as you mentioned).
If you do the same for a location ahead of the LE, you will get a net upwash, and inverse for TE. This seems intuitive: if you zoom out and treat the whole wing as one single circulation, the circulation should be clock-wise (assume air comes from the left), with air moving upward to the left and moving downward to the right.
There are more complex models, like vortex lattice, vortex panels, etc. But this theory is fairly intuitive and introduces the induced drag. By the way, induced drag is a finite span phenomenon. If you have an infinite span, you fully recover the 2D lift coefficient and the induced drag disappears, even though the vortices remain.

Although tip vortices are usually thought of as a little horizontal tornado emitting from the trailing edge wing tip, it's actually a large flow field extending ahead and behind, and also beyond, the tip.  Both the upwash and the tip circulation flow starts well ahead of the wing.  It's just most intense near the tip's surface toward the back.
You can see this if you examine a winglet standing under it and notice that its airfoil's chordline is pointed outboard a couple degrees to maximize the AOA in the tip circulation, because the circulation flow field was started well out front and the flow is already curving inboard at the winglet's leading edge.

So farther inboard, it's just the upwash "bow wave" of the moving body of the wing ahead of it, but toward the tip the upwash is already starting to circulate around to the top, while it's still upwash, so the net angle is higher.


text:
How practical is laser designation against helicopters?
Laser designation is a fairly common method for guiding missiles and precision bombs, but how does it perform against aerial targets?
Given the ranges and kinematics involved, this method could be too challenging. I know the effectiveness is also dependent on atmospheric conditions, but what are the other limiting factors?

A F-15 shot down an airborne Hind: https://theaviationgeekclub.com/that-time-an-f-15e-shot-down-an-iraqi-mi-24-attack-helicopter-with-a-2000lb-laser-guided-bomb/

The 9K121 Vikhr beam riding missiles, usually employed against armored targets, have a fusing mode specifically against airborne targets and a secondary warhead to increase their effectiveness in this mode, meaning there is some doctrinal belief in their effectiveness in this role.

It might not be as hard as you think, especially if the helicopter is heading your direction (e.g. used as a defensive weapon, not interception weapon). All you need to do is to aim the laser on the target. It's more or less like aiming a rifle at a fast moving car. Compared with anti-helicopter rifles (e.g. M82A2), laser guided missiles looks quite usable.

There are a number of laser-guided ground-to-air systems, and a number of those have been adapted to air-to-air and air-to-ground as well. The key is the speed of the missile, not so much the guidance system itself. An example is Starstreak.
Lasers are well focused, but not perfectly, and are easy enough to defocus to any required degree. For this role, making the beam large enough to cover potential movement of the target during the tracking cycle (often manual) while still having enough energy fall on the target for the tracker to seek is pretty much all you need. Modern solid-state lasers make this relatively easy. However, unless you have a seeker in the munition, as is the case in a laser-guided bomb, then the cone-shaped beam means decreasing accuracy with range.
Starstreak takes another approach. It projects a 2D array of dots in space modulated so the receivers on the darts can tell where they are in the array. When used with the shoulder launcher, the operator keeps the beam on the target optically and the darts keep themselves in the center.
When combined with the optional launcher platform, a tracking camera is added. This receives the reflected laser signal from the target and uses the same modulation to determine how closely the array is to the center of the target. It uses this to automatically track the target, and thus the missiles get the benefit of semi-automatic tracking without having to carry the seeker on their nose.

The Starstreak relies on beam riding, not target illumination. If painting is feasible, considering the difficulty of evasion, why are there no SAM systems that utilize it? Could APKWS or Hellfire engage a rotary-wing aircraft painted by a SOFLAM or the likes?

Starstreak Lightweight Multiple Launcher has IR search and laser (lidar) tracking in addition to the guidance beam. There is an air-to-air version under development as well. SAM systems do not use it because they generally don't have the Starstreak's mission - which is to engage helicopters flying behind trees. If you are looking to shoot down aircraft "in the clear", a radar system is dramatically more economical and long-legged. As to Hellfire, I believe it HAS engaged a helo, but I'd have to look that up.

And we should not forget the RBS 70 as well, although that is purely beam-riding.


text:
What Microcontroller(s) is/are used inside the ECU of a CF6-80C2 engine?
The question came up at work today and no one knew the answer. Did some digging online and could not find anything definitive.

That information could be a trade secret, therefore not likely to be public.

Don't know for sure so can't post this as an answer, but given the age of the design I'd say the probability is high it uses an Intel 8086 microprocessor.  There is usually a command and a monitor channel running together in these sorts of units, the monitor channel mimicking everything the command one does to detect anomalies, and often monitor channel might use a Motorola equivalent like a 68000 to mitigate the risks of a processor architecture related defect causing both channels to see the same flaw and not see anything wrong.


text:
What does X on a private runway mean?
Does an X at the ends of a private runway mean it's inoperable or does it mean its restricted to the public?

Possible duplicate of Why is a large flashing "X" placed on a closed runway at Reagan National Airport?

@AEhere I think OP is aware that X means "closed" but would like to know if the reason of the closing can be inferred for a private runway.

@QuentinH That is primarily-opinion based as we do not read minds. There is a question to be asked, however, about the suitability and/or legality of closing a working (and potentially active) runway via an X marker, but that is too removed from the OPs wording.

I agree. Not a duplicate though

Related

One of the reasons why a private runway owner would "X" it would be to indicate that the runway is not in good condition or closed. This can be for a genuine concern or to avoid liability concerns.

Or just because they don't want you to land on their field.

@QuentinH Yes, this was addressed in my answer (Closed)


text:
Flight engineers - commercial aviation
Apologies for being so broad in the title but this will consist of a couple of questions in one.
As I'm sure most of use are aware modern commercial aircraft being manufactured no longer incorporate a flight engineer, this role - as far as I know - has been taken over by onboard systems monitored by the pilots.
Obviously there are a number of airlines out there flying older aircraft that do require a flight engineer, which airlines still actively hire F/E's? Do airlines still train people for this position or purely rely on hiring people already trained? I'm thinking more along the thought process here of 'When the older F/E's retire, how are they replaced?'
Finally, what happened to the F/E's when this job was phased out? Were they retrained as pilots? Merged into an office job? Flat out dropped from the company?

Related: As 727s retire, so do their flight engineers and Flight Engineer - A Dying Breed

Asking multiple questions in a single question is considered bad form on the Stack Exchange network. Also, at least "which airlines still actively hire flight engineers?" is a list question, which in itself is a poor fit for Stack Exchange as it is difficult to provide a single, authoritative answer to it (never mind that the answer will change over time). At the very least, consider asking each subquestion separately; even though they all deal with the flight engineer role, they are separate questions and should be treated as such.

@MichaelKjörling Once the list question has been discarded, I think there are two questions here: how do airlines currently meet their flight engineer needs? and what happened to flight engineers as that role disappeared? Most of the subsidiary questions (Did they become pilots? Did they get office jobs? Were they just made redundant?) are essentially yes-no questions that couldn't stand on their own anyway.

@DavidRicherby I see basically "which airlines still actively hire F/Es?", "do airlines train F/Es?", and "what happened to past F/Es?". There are at least two separate questions there, "how do airlines gain flight engineer expertise currently?" and "as the flight engineer role became less and less needed, how did airlines respond to that?", which IMO are different questions. (That said, I'm not an aviation expert, just a fairly-long-time SE member, so maybe I'm in the wrong here. That's why I wrote "consider" in my original comment.)

@RedGrittyBrick Many thanks for your great reply. That information pretty much covered everything I was asking. If you could post your answer then I will mark it as the correct one.

I heard of an engineer from New Zealand taking a course in USA back in 2006 to gain the F/E qualification on a Boeing 727, so apparently it can still be done. I imagine the flight training centres in USA which specialise in flight simulator based type certs must still teach it.

SIR PIA ARE STILL OPERATING THE B747 367 AND OPERATING FLIGHT ENGINEERS ON THESE TYPES. FROM RAYHAN AHMED (CARGO AIRCRAFT LOADER)

Couple years Pakistan airlines came in with a B747 367 AP BGG (heathrow)once we pushing the aiecraft it developed engine problems. I thing no 3 or 2 engine was not starting up due to fuel pump defect . it took many minutes for the flight engineer to start the engine. we thought the capt might bring the aircraft back on stand but due the flight engineer they managed to relight the engine and taxy off. Having a human being attending to a defect is much better then a computer like what they have on - 400. From RAYHAN Ahmed (cargo aircraft loader)

Otherwise without the flight engineer at time we would have to offload all cargo pallets containers and maybe it would stayed overnight or the hanger. From RAYHAN Ahmed (cargo aircraft loader)

Note PIA are going to retire there flight engineers on final phase of the -367 which going to replaced by 777.

This will be in 2015. This is a real shame.


text:
Why is Tromsø Airport's runway 01 referred to as 36 in the sign before the runway?
Runway designations can change due to the changing direction of the magnetic north. Using Google Earth, we can verify that Tromsø Airport's runway 01 has not changed its designation since at least 2005. Nevertheless, as we can see at the start of this video the sign just before the runway refers to the runway as runway 36. Does the 36 on the sign refer to the true direction while the number painted on the runway is the compass direction?

Also, closely related: How do airports determine when they should change their runway numbers?

The runway number is changed to 18/36 due to the change of magnetic variation. The image on Google Earth must be outdated. This is from my current (AIRAC 2004) Jeppesen airport chart of ENTC (Tromso, Norway):

The video you linked was uploaded in March 2020, were the runway number had already been changed. I do not know how old the Google Earth image is. According to this thread on a simulation forum, the runway number was changed with AIRAC 1910, which came into effect on 12 Sep 2019.
By the way, both the numbers on the runway and the sign refer the magnetic heading. For more details, see How are runways numbered?.

If I'm not misunderstanding it, Google Map stated that the imagery data is from 2020, but the paint on the runway is still 01/19.

@AndrewT. No, that 2020 is the copyright information, not the year when the image was taken. See this Google support question: "You can only check the age of satellite imagery using Google Earth Pro".


text:
Can a harbor freight predator 212 cc engine be enough for an ultralight
Hi I am planing on building a budget ultralight and I was wondering if a 6.5 hp 212 cc predator motor will be enough and if so what size prop should I use.
A little background, it is going to be a wood and foam plane with fiberglass covering I am going to have a few bits and pieces of aluminum for strength. I don’t have a plan yet because I am going to plan it around the engine. I am shooting for 160 pounds dry. Thank you in advance.

Even paramotors, that don't have any airframe (really just a propeller you strap on your back with some safety cage) use 20+ HP motors. 6.5 HP is not going to get you + airframe + fuel airborne without some very fancy modifications.

6.5 hp is enough to get a person off the ground - but it would require an extremely aerodynamically- and structurally- efficient airframe to do so. A first-time builder without experience or training is not going to build such an airframe on the cheap from wood and foam and fiberglass with "bits and pieces of aluminum for strength". Either the wings will break and you'll kill yourself, or more likely it just won't get airborne.
You should scrap this idea and instead find a good set of plans and follow them exactly.

There's a good reason most ultralights, even in the early days, used 3-5 times this much power: because 20-30 hp is the minimum needed to give reasonably safe takeoff and climb performance.  You will not meet your goal of 160 lb dry if you provide enough wing area to lift that plus a pilot; even a fabric hang glider type ultralight, by the time it has a seat, landing gear, and engine mount, will be at or above this dry weight -- and even with your goal dry weight, add a pilot and five gallons of fuel, and you'll need a long takeoff roll to reach a safe takeoff speed.
You've got around 90 lb to spare under the ultralight limit (assuming you're in the USA): I'd recommend allowing half of that for overruns in the airframe, and putting the other half into a larger, more capable, likely more reliable power plant.  Get something that's actually intended for ultralights -- you'll get more power per pound, which makes everything else better, and you'll decrease the likelihood of a failure at a critical time (like at low altitude after takeoff).


text:
Does anyone have free plans for a ultralight?
I am looking for free ultralight plans for a beginner. I have been wanting to fly but I don’t want to spend thousands on a license so ultralight is the next best thing.

David, if you don't mind me asking, exactly how old are you?

Learn to fly first.  Don't teach yourself.  In addition to ultralight lessons, you might seriouly consider taking some hang gliding lessons, either by foot-launch or aerotow.  It will be a great experience and you will learn some interesting things about flying and it will be easier to do while you are young and fit than later in life.  Just be sure to find some place where they don't make you buy all the gear, unless you are sure you want to make hang gliding your main hobby.

Your best bet is a Sandlin Goat, with free plans for download and quite a few built.  It's a soaring glider but some have added motors, although that eats into the useful load.  A motor would have to be a 2 stroke to have adequate power for the weight, (you want about 25 hp) and it would be best to use an off the shelf paramotor engine.  I would just purchase an old used paramotor cage/motor to get the motor/reduction drive/prop and rebuild the motor.  A goat itself is still going to cost you at least 5k to build from scratch, probably at least half that in the cost of fabric covering process, before the motor.  Then of course you have to engineer a way to attach the motor with adequate structure that can handle several Gs, doesn't screw up the weight and balance, etc etc etc.
Aviation, even at that level, ain't cheap.
But before all that, you have to get training to avoid killing yourself, so save up a couple grand first and go to an ultralight flying school for lessons.

+1 The third paragraph is the most important.  It's easy enough to kill yourself in an aircraft when you've lots of experience and training, with neither it is pretty much a certainty


text:
How can the effects of hot corrosion and salt deposits from sea water spray in gas turbine engines be negated/prevented?
In aircraft operating in a maritime environment, corrosive effects from salt water are common. Are there any specific coatings used in engines for such aircraft? If so, which are they? How frequently should they be washed?
I've looked for information on this but can't find any specific answer.

Not aware of any anti-corrosion coatings used in engines.  The best mitigation is to shoot fresh water into the inlet with it running at the end of the day.  Seaplanes operating off salt water will generally get a fresh water hose down at the end of the flying day whenever practical.

There are different metallurgical mixes that react more or less violently with seawater, and the Russians were, and are, very good at that, not least Beriev, so their engines are made to operate near seawater, thus need to have different metals used internally, compared to ordinary engines.
As we all know a lot differs between an RAF Harrier and an RNAF Harrier, not least due to their different operating milieu, like the plating used internally. And I guess the same applies to various models of the F-35?!
I do not know anything in detail, but might be available. Magnesium is avoided in a salty context, lithium-aluminum works well (if my memory serves me right), as does Alclad.


text:
Why doesn't a wing in supersonic flow produce a shock that results in flow parallel to the leading edge?
When we look at a wedge in supersonic flow, a shock emanates from the leading edge such that the flow after the shock is turned and is parallel to the airfoil surface.

The shock angle is based on the flow turning angle (wedge angle) and can be found using the chart below:

I was wondering why a wing in supersonic flow would not have the same behavior (see image below). That is, why wouldn't there be a shock emanating from the apex of the wing (which depends on the sweep angle much like the wedge angle ($ \theta $)? If this were to occur, there would be no component of the flow perpendicular to the wing (as the shock would turn the flow parallel to the surface). I know this doesn't occur physically because there is always a component of flow perpendicular to the wing, but this confused me because the wing seems exactly like a wedge (if we look at a wedge from the top instead of the side).

The two-dimensional flows, in section and plan, are just approximations or simplifications which apply under certain conditions. This situation can be easier to understand as a three-dimensional flow, and perhaps also from the reference frame of the air. When the wing approaches, the resulting pressure gradients make the air move out its way. Because the wing is relatively thin, the vertical gradient is much stronger than the horizontal one. So the air tends to move that way.
However some sideways flow does occur. Provided the shock cone remains in front of the wing, air has time to move sideways as well as up. Due to the sweep of the wing, there is a difference in pressures on either side of any given point. These make the air start to flow out from under the leading edge, curl up round it further out, get drawn back in again and then part way back over the wing get pushed out again. This sideways flow pattern tends to affect the pressure differences in turn and hence the performance of the wing. On a very sharply-swept delta it can even dominate the airflow. However for a more conventional swept wing, devices such as wing fences, leading-edge slots and dog-teeth are sometimes incorporated to reduce it.

Air needs to make way for the approaching airplane. For that, it takes the route of least resistance, which is above and below the wing. This is shown in your first picture which is valid for a supersonic leading edge. Your top view shows a different cross section through a three-dimensional shock at a much lower speed. The vector M$_2$ is wrong and does not correspond to what is really happening.
If X is in flow direction, Y is sideways and Z is down, picture 1 shows a cut in the X-Z plane for a supersonic leading edge. Picture 3 shows a cut in the X-Y plane for a subsonic leading edge. Here, air does not flow sideways but continues past the shock along the direction of the vector M$_1$.


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Why do shockwaves progressively move aft as speed increses?
As an airfoil moves from subsonic to supersonic, the shocks that develop over the surface move progressively toward the trailing edge. What is the physical explanation for this behavior?
As a side question, what determines where the shock (and hence location of minimum pressure) is located?

The answer is deeply rooted in the theory of compressible fluid dynamics, so for a fully satisfactory take on the matter you might want to refer to a textbook on that topic (such as Thompson or Shapiro). I will try to give a qualitative explanation.
For our discussion here, let's consider a shockwave as the way to abruptly slow down a supersonic flow and recover pressure; abruptly, because, since disturbances cannot propagate upstream in a supersonic flow, the flow has no way to "know" that the disturbance is coming and it isn't able to "smoothly" accomodate it (as it happens in subsonic flow). A disturbance here is, for example, the presence of the airfoil and its effect on the velocity and pressure fields.
Let's have a look at your picture. In a) the flow is subsonic everywhere and we have no particular issues with that (although we might want to consider some compressibility corrections).
Now, the flow accelerates over first portion of the upper surface of the airfoil, so the speed of the flow there is greater than the speed of the free stream flow (and greater than the speed on the lower surface, so we can have lift).
We can see that for free stream Mach numbers $M_\infty$ (ie flight speeds) close to 1 the flow over the upper surface can actually reach local Mach number $M$ of 1 and even above, because of that acceleration.
This is exactly what has happened in b): here $M_\infty$ is greater than the critical value (usually given as 0.8) and there is a supersonic bubble right there in the middle of the upper surface.
The flow downstream, however, is still subsonic, therefore it has to slow from the supersonic condition it reached and to do so a shockwave appears. This shockwave usually interacts with the boundary layer, adding a very high adverse pressure gradient to the one already experienced towards the trailing edge, so the boundary layer will most likely separate, leading to stall.
As we increase the flight speed, more and more of the flow over the airfoil becomes supersonic, also on the lower side, therefore the shockwaves "travel" downstream until, close to $M_\infty = 1$, the whole flow on the airfoil is supersonic, as in d).
As you can see from c), the shockwave on the lower side actually overtakes the one on the upper side. This is because the lower side is usually "flatter", so the shockwave appears later (ie at higher $M_\infty$) since it experience less acceleration (which is due to the curvature), but it also experience less deceleration towards the trailing edge and the local flow will be supersonic over a longer distance along the lower surface.
For even higher flight speeds, a detached shockwave appears in front of the airfoil: the purpose of this wave is to "turn" the flow around the airfoil. As the flow is supersonic, this turning of the flow can't happen "smoothly", but like we said it is the shockwave that abruptly accomplishes it.
The flow behind this curved shock is still supersonic (except for a region right around the nose of the airfoil) so later downstream we once again find the shockwaves which help the flow go back to its undisturbed free stream conditions.
The detached wave significantly increases drag; in some application this is optimal (eg in the atmospheric re-entry of a spacecraft), but it is detrimental for supersonic flight. This is why supersonic airfoils are designed to have a sharp trailing edge, so that the turning of the flow can be accomplished by oblique shocks attached to the edge, resulting in less drag.
As to answer your last question: where does the shockwave position itself? The answer lies in the boundary conditions, that is to say the conditions that have to be restored downstream of the airfoil: these will be the free stream speed and pressure.
Without going into much details about the variation of pressure with velocity in compressible flows, let's just say that the shock position is such that afterwards the flow has enough "space" left to recover subsonically. So, say you have to recover to a pressure of 1 to the freestream pressure of 10 (random numbers with no meaning) at a certain distance downstream from the airfoil: if the shockwave increases the pressure to 6, it will be positioned in such a way that after the shock (where the flow is subsonic) there is enough space for the pressure to increase "smoothly" (subsonically) by 4 and restore freestream conditions at that distance.
The increase of pressure across a shockwave is given by the shock strenght and can be seen as proportional to the Mach number, so at higher Mach the shock is stronger, the pressure recovery is higher and the shock will position further downstream, since the flow will have less pressure to recover subsonically and can do so over a "shorter" distance.

the crux of your answer of why the shocks move toward the trailing edge relies on the fact that "more and more of the flow over the airfoil becomes supersonic," as M increases. How can you justify this physically? Why does the extent of supersonic flow not stay the amount it is in figure b) and increase to the amount it is in c)?

With a higher flight velocity (still subsonic), the supersonic pocket on the upper side of the wing starts earlier and takes longer until it collapses into the shock. That is quite normal, as the speed over the whole upper wing is increased with increasing flight speed.
Think of what happens at the extreme case of accelerating flight speed towards Mach 1: Now the shock will move towards the trailing edge with increasing Mach number and will stay there once the flight speed is above Mach 1 (NOTE: This is valid only for straight wings; correct the Mach number upwards for sweep effects).
Things become nasty when the shock interacts with lift creation and starts to fluctuate back and forth - this creates a buzzing sound and might even cause flutter. The mechanism is that the stronger shock causes flow separation and lift loss, which in turn shrinks the shock, which allows lift to recover so the shock grows again. With the growth, the shock moves back and jumps forward when it becomes weaker. This back-and-forth jumping of the shock creates not only a fluctuation of lift, but also of pitching moment.


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How does wing shape impact shock structure?
I am trying to understand what the 3D shock structure would be for a wing in supersonic flow. Suppose we have a wing which is made of double-wedge airfoil cross-sections.
If we view the shock structure over the airfoil (side view), it would look like so:

Now, if we look at the top view, we would see the following structure (top view):

Questions:

Is the shock seen in the side view (airfoil shock structure) related to the one seen in the top view?
How would the shock structure change if the airfoil is kept the same but the wing is swept more?
(If you can show some 3D plots for the shock structure, it would really be helpful!)
Can you explain what determines the shock angles in this picture (is it sweep or airfoil shape or both - and which is more dominant)?

Is the shock seen in the side view (airfoil shock structure) related to the one seen in the top view?

No, not at all. The first picture (side view airfoil) shows an unswept wing in supersonic flow and is not very precise. The shock angles of the LE shock and the TE shock should be identical.
The second picture (top view wing) shows a swept wing at bit above the speed of sound (almost half the speed in the first picture). Here, the flow pattern around the airfoil is fully subsonic (that's why the wing is swept!) and only its most forward tip will cause a shock which is almost straight because flow speed is only a bit higher than the speed of sound.

How would the shock structure change if the airfoil is kept the same but the wing is swept more? (If you can show some 3D plots for the shock structure, it would really be helpful!)

That depends on the ratio between speed and sweep angle. As long as the sweep angle exceeds the shock angle, flow around the airfoil behaves like in the  subsonic case with a stagnation line, flow around the round leading edge and no bow shock or expansion fans. When the shock angle is greater, the airfoil will show a flow pattern like in the top picture. Sorry, no fancy 3D plots!

Can you explain what determines the shock angles in this picture (is it sweep or airfoil shape or both - and which is more dominant)?

The first and foremost factor for the shock angle is speed. Mach 1 will cause a straight shock and higher speeds will bend the shock backwards into an oblique shock with a shock angle $\varphi$ according to the equation $$\varphi = arccos\left(\frac{1}{Ma}\right)$$
This angle will become smaller with increasing bluntness of the body causing it.

Shock angle diagram (from Wikipedia, source). The caption is: This chart shows the oblique shock angle, β, as a function of the corner angle, θ, for a few constant M1 lines. The red line separates the strong and weak solutions. The blue line represents the point when the downstream Mach number becomes sonic. The chart […] is valid for an ideal diatomic gas.
Explanation: M1 = Mach number ahead of the shock, M2 = Mach number past the shock. β is the half cone angle such that β + $\varphi$ = 90°.
Consequently, the angle of the shocks emanating from that T-38 show that it flies at around Mach 1.09. In Schlieren images like this darker areas indicate higher density and lighter ones lower density. The two dark lines coming from the fuselage tip and engine exhausts are the two compression shocks which are audible as a double boom when a supersonic aircraft passes overhead.

If we look at the picture of the wing from the top view, we see that the sweep causes the supersonic free stream to turn through a finite angle (specifically, the sweep angle). Since a supersonic flow turned in on itself generates a shock wave, can you explain why the sweep angle would not change the shock which emanates from the leading edge of the wing? Basically, why is the shock angle determined by the formula you gave (for Phi) (which is indep on sweep angle) and not by the graph you showed?

@NickHill "we see that the sweep causes the supersonic free stream to turn through a finite angle" – do we? That drawing isn't very helpful. The Schlieren photo with the T-38 is much better and here you see the shock angle and can use it to directly determine speed. The bow shock has a slightly rounded tip. The local angle close to the nose is what can be read from the diagram, the far-field angle is given in the formula.

Suppose we have a flying wing with a certain sweep angle (lambda). Would the shock emanating from the apex depend on lambda at all? I'm just a little confused by why the shock angle would not be changing if the flow is changing direction starting at the apex of the wing and is being turned in on itself. Should this flow turning not create a shock wave that depends on lambda?

@NickHill: How should that work? Pressure changes travel with the speed of sound, so any change to the air cannot happen ahead of the shock front, independent of what comes after the apex as long as Lambda is larger than the Mach cone. There simply is no way the wing can change the shock angle as long as sweep prevents it from doing so.

I clarified my question further here:  https://aviation.stackexchange.com/questions/79593/why-doesnt-a-wing-in-supersonic-flow-produce-a-shock-that-results-in-flow-paral


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What is the physical-based/theoretical reason why supersonic area ruling works?
Supersonic area ruling is used to reduce the drag rise that occurs as aircraft pass through the transonic regime. While it has been validated experimentally, what is the theoretical/physics-based reasoning why area ruling reduces the drag?
I know that the following answer suggests that it is simply a "rule," but could someone shed more light into WHY the area ruling technique works? (What is the "area rule"?)


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Why do swept wings decrease the drag increase that occurs in transonic flows?
Swept wings are used in aircraft that fly in the transonic regime because they 1) delay the critical Mach number 2) reduce the drag increase that occurs (smaller drag divergence).
I am trying to understand physically why a lower drag is observed. A common explanation is that only a component of the free stream is perturbed by the wing (see: Why are swept wings better for breaking the sound barrier?), but this does not explain why there is a lower drag when shocks do occur over the wing.
More specifically, I understand why the curve below shifts to the right as sweep increases, but want to know why the drag is not as high.

Source: https://history.nasa.gov/SP-468/ch10-4.htm

Shock waves form where air can no longer "get out of the way" and begins to compress.
Sweeping the wing effectively increases the chord and reduces the rate of wing thickness increases as airflow moves across it, essentially making the body longer and more "streamlined". Instead of running into a steep hill and compressing, it can flow over a more gently sloping barrier with greater ease.
What is most interesting is the decreasing drag of the 47 degree sweep at high subsonic.  As sweep increases, airflow over the wing may transition from the classical front to back flow seen in straight wings to the deflected "spanwise" airflow of swept wings to the rolling vortices seen in highly swept "slender delta" wings.
At some combination of speed and sweep, airflow will roll up into a vortex rather than washing off the back or end of the wing.  It may be that, at higher subsonic speeds,  with greater sweep, the vortices roll into the back of the wing and push the aircraft forwards.
It has been thought the same thing occurs underneath thin undercambered wings, which may help account for some of their "magical" slow flight properties.  Definitely worth studying.
As far as 11, and 35 degrees of sweep, they follow predictable patterns based on compression effects at transsonic speeds, as does the Area Rule.
Are swept wings better for breaking the sound barrier?  Not really, as the X-1 proved.  But they do significantly delay transsonic drag at high subsonic speeds.

By effectively i creasing the chord, are you reffering to spanwise flow?

That as well, but if you start with a straight wing and sweep it, relative to the wind flow (even if it is not deflected) the chord (distance from front to back) is greater.

Yes, but the profile would be different. Say you want to use naca xyz, because it is suitable for flight characteristics you want, you cannot use the profile perpendicular to the span. It would not behave as naca xyz because the airflow is coming at an angle to this profile. You cannot compare profiles like that.

It would be good to see how 2D polars changed adjusted for sweep, both computer and wind tunnel data.  Spanwise flow will also cause "variation in profile" towards the wing tips.  I'm really interested in the conditions where spanwise flow begins to roll into vorticies as sweep increases.

Me too. I'm not an aerodynamicist (?), but I bet the sweep enhances spanwise flow, and this gets stronger as the plane goes faster, thus "slendering" the wing profile as seen by the airflow.

Let's consider an infinite swept wing with a sweep angle $\Lambda$. We can divide the freestream velocity ($V_\infty$) into two components: the component perpendicular to the sweep ($V_{\perp}$) and the component parallel to the sweep ($V_{||}$). Similarly, we will divide the exo-boundary layer velocity field around the wing ($V_e$) into the perpendicular component ($u_e$) and parallel component ($w_e$).

Graph cited from Drela, Flight Vehicle Aerodynamics.
Due to infinite span, the flow quantities along the swept coordinate ($z$) must be invariant. Therefore, the flow parallel to the wing sweep ($w_e$) must also be invariant and equal to the freestream component ($V_{||}$) everywhere.
Assuming that the flow outside of the boundary layer contains only weak shocks and is therefore isentropic (except at the shock), we can use the isentropic relationship to relate pressure field and velocity field. Here, everything with subscript $_e$ denotes flow quantities outside of the boundary layer:
$$p_e(x)=p_\infty \left(\frac{h_e}{h_\infty} \right)^{\gamma/(\gamma-1)}=p_\infty \left[ 1 + \frac{\gamma-1}{2}M_\perp^2 \left(1-\frac{u_e^2}{V_\perp^2} \right)\right]^{\gamma/(\gamma-1)}$$
where $h$ is enthalpy, $p$ is pressure, $M$ is Mach number, $\gamma$ is specific heats ratio.
Notice that the flow parallel to the wing sweep has no bearing on the pressure field; it's as if each wing section only sees the perpendicular flow. The same goes for effective 2D Mach ($M_\perp$), which is a net reduction from the freestream Mach ($M$) by a factor of $\cos\Lambda$. Since a reduction in Mach reduces the normal shock strength in the 2D section, or even removes the shock altogether, it leads to a reduction in wave drag compared to no sweep.
Below is the 2D drag curve vs. Mach for NACA 0012 from Euler solution. Local shock is obvious starting 2D free-stream Mach 0.65. This goes to show that decreasing effective 2D Mach lowers the overall wing drag via sweeping.

Graph cited from http://aerodesign.stanford.edu/aircraftdesign/drag/dragrise.html.
In a finite span wing, there will be gradient in $w_e$ that will lead to flow variation along the span. However, the principle of wave drag reduction through wing sweep readily applies.

Could you please provide the reference to the formula of $p_e$?

@Hans See https://www.grc.nasa.gov/www/k-12/airplane/isentrop.html relation 4. Under calorically perfect gas, Cp is a constant, thereby giving the relationship in my post.

Thank you. It would be better to add the reference to the main text of your answer below the equation.

@JZYL At some point, Mcos(Lambda) is going to be such that a shockwave occurs over the wing. Your explanation does explain why the critical Mach number increases, but can you clarify how the actual drag decreases when the shock DOES occur? In particular, the shock strength is reduced at a given free stream Mach number, but how does this explain why the curve in my original question has a smaller peak (albeit at a higher Mach number)?

@Nick Are you asking why the maximum wave drag decreases as a function of sweep? 2D shock strength is a function of Mach number.


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What is the impact of camber on CLmax?
I was reading Aerodynamics for Engineering Students by Houghton and saw the following graph:

A lot of other sources say that increasing camber increases CLmax, like the graph below.

I am a little confused by what camber actually does to CLmax? An explanation which focuses on the physics would be very helpful!

I think the bottom one is correct from reading papers which measure, or at least calculate it. but I don't know why the textbook would be wrong. Papers such as this: https://arc.aiaa.org/doi/pdf/10.2514/1.C034415 If the top were true, then only ailerson would have any use and flaps would be useless.

Camber directly affects the amount of circulation an airfoil can generate, adding camber will always generate a higher zero alpha lift coefficient (intersection of the lift coefficient curve with the x=0 axis) up until a point that there is too much camber to sustain attachment.
This leads directly to the question asked - does camber increase CL_max? For any given AOA, there is an ideal suction side geometry that generates a pressure distribution that is always on the edge of separating. Liebeck defined this ideal shape but only for laminar flow, for which the boundary layer equations are closed form, this was all based on the work of Stratford who came up with the ideal pressure recovery. Having a surface that is about to seperate everywhere can even be true for a symmetrical section at some inclination, although it's not going to look like a NACA 0012/15 or anything familiar. The effect of the pressure surface on the amount of circulation generated, is much more insignificant than the suction surface, but it is not zero. Therefore, a cambered airfoil will make slightly more load at max CL than a symmetric section.
In reality, there is no magic button to increase, decrease camber. Airfoils are complex geometries that cannot be driven by one parameter without affecting everything else. Adding camber by 'bending' the airfoil somewhere near the front of the chord will give some more circulation, but also probably increase the departure angle, and create a risk of separation towards the trailing edge where the pressure recovery is too steep for the boundary layer to cope with.
Unfortunately there is no easy +camber = +max lift coefficient relationship, but for a cambered and non-cambered airfoil designed to operate on that separation limit, the cambered airfoil will generate slightly higher max lift.

Could you elaborate on your 2nd paragraph. I am still unclear on how you came to the conclusion that "camber will make slightly more load" based on what you said about the suction/pressure surface.

Sure @NickHill, I have added a bit more context about the "ideal suction surface". Once that concept is accepted, the question really becomes "can I add more circulation to the pressure side of the airfoil by changing it from symmetric", to which the answer is yes. But the answer to "can I add MUCH more circulation to the pressure side of the airfoil by changing it from symmetric" is unfortunately no.

@NickHill does this answer your question?

In general, boundary layer development is key to understanding the role of airfoil geometry on performance metrics such as CLmax. In this case, with increasing AoA, positive camber creates a reduced adverse pressure gradient near the leading edge thus delaying the onset of massive boundary layer separation at stall angles relative to a symmetric airfoil. So, with positive camber, CLmax could be higher since massive flow separation occurs at a higher AoA.
Of course, you could use camber to reduce CLmax relative to symmetric airfoils.

Your answer seems to suggest that CLmax occur at a higher angle of attack. If you see the second graph in the figure above, camber actually decreases the stall angle

@Nick Hill  lowering the leading edge decreases stall angle AOA as measured by the chord because the bottom camber is removed (as explained in paragraph 2 of my answer).  But there is no magic here, the Clift/Cdrag is lower for the cambered wing, requiring more thrust for higher Lift.  This is why the Fiesler Storch needed a 240 HP engine to do its legendary feats.

Camber is measured by taking the midpoint between the top and bottom of the wing and comparing it to the chord line, which runs from the leading edge to the trailing edge.
The top graph simply shows that bending the leading edge down, as slats do, will begin to generate lift even if the wing is at zero AOA, because the top of the wing now has a greater
curvature than the bottom.  Notice a symmetrical wing (no camber) at zero AOA will have its upper and lower "lift" cancel out, and must be at a positive AOA to generate lift.
The undercambered wing in the second graph will produce a greater Coefficient of Lift because of it's ability to "trap" more air underneath the wing and force it downwards.
But one should be aware of the drag penalties of this type of wing, requiring much more thrust generate the same amount of lift as the symmetrical airfoil moving faster.  This is why cambering (with slats and flaps) is generally not utilized in cruise.

How does this explain why CLmax increases with camber?

See paragraph 2 and 3.

"trap" is a very dangerous term to use. All but a tiny amount of the load is generated from the suction side of the wing, especially in high lift configurations. Looking at chordwise pressure distributions of inclined sections vs cambered sections, you can see that the pressure side pressure distribution changes very little.

@Stuart Buckingham the most efficient lift comes from the suction side, but a significant portion comes from the pressure side by simple action/reaction of air being deflected downwards.  Notice top lift on super critical wings is verboten because of Mach effects; there is still enough underneath to lift an airliner (if it is going fast enough).

The 'air deflected downwards' explanation of lift can be very misleading. An inclined flat plate will deflect air downwards, but does not provide an efficient mechanism for creating lift. Lift is better thought of as a function of circulation around an airfoil instead.
Looking at the pressure distributions on a section, particularly at large alpha, the CP on the pressure side is very close to 0, meaning it is not contributing any load, whereas the suction side can be up to CP -5

@Stuart Buckingham I suggest you consider water skiing.  In the air, pressure increase may be small, but the amount of air being deflected is large and continuous. (Water skiing is bottom lift under slightly more favorable circumstances, but it works with air if you go fast enough.)

(Or if the skis are larger and the air is dense enough).

@RobertDiGiovanni that is very true, the bottom surface of the ski is obviously generating enough load to hold the skier up, however, you are now looking at essentially a 'one-sided' surface, as the flow across the top of the ski is a completely different phase. Comparing pressures on a non-opposed surfaces then becomes even more confusing (Doug McLean has a fantastic explanation for this in his book "Understanding Aerodynamics: Arguing from the Real Physics") - for instance you cannot compare the drag on the front half and back half of a vehicle.

@Stuart Buckingham the real eye opener for me was when I saw a lift vs AOA graph all the way out to 90 degrees.  At 45 the inclined plane produces similar lift to just before "stall" AOA, albeit with hugely more drag. Agree with circulation and top lift, but think downwash from bottom inclined plane helps accelerate the air flow over the top within the nonstalled regime.

@RobertDiGiovanni very true regarding lift vs AOA for thin sections or flat plates. Not quite sure what you mean about the pressure surface accelerating the flow over the suction surface.

@Stuart Buckingham airflow deflected downwards will create lower pressure behind the wing which will help accelerate airflow rearwards.  At non-stall AOA, this gives top lift a little extra with very little penalty in drag, resulting in that lift "spike" one sees on the lift vs AOA graph.  Just my thoughts.


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Why does lift curve slope increase when Fowler Flaps are deployed?
Below is an image for a Fowler Flap:

Notice that as the flap is deployed, both the camber and the area increase. This change in wing shape modifies the lift curve as follows:

(source: Development of a Fowler Flap System for High Performance General Aviation - Wentz (NASA))
This behavior can be isolated and viewed in the simpler figure below:

Note that the lift curve slope has increased.
Here is my analysis:
If Fowler Flap is deployed, the planform area increases.
If the planform area increases, the aspect ratio decreases (since wing span is unchanged).
If AR decreases, then the lift curve slope should decrease (see How does the aspect ratio of a wing impact its lift? for details).
So where in my analysis do I go wrong? I know the lift curve slope should increase, then why does my analysis indicate otherwise?

Note that your plots show lift coefficient over angle of attack. This lift coefficient is referenced to a reference area which must be defined somehow. It is customary to use the projection of the clean wing area in the x-y plane for all flap settings in order to keep coefficients comparable. The lift curve slope increases because the real wing area increases while the reference area remains unchanged. The dashed line in the lower plot shows the isolated effect of camber increase while the upper solid line combines both effects (area and camber increase).